Is there any closed formula for such integral $$ \int_{-1}^1 x^i T_n(x) dx? $$ here $i$ is an integer numbers.

For $n=2$ I have found that $$ \int_{-1}^1 x^i T_2(x) dx=\begin{cases}0,i=2k+1 \\ {\frac {4\,k-2}{ \left( 2\,k+2 \right) ^{2}-1}}, i=2k, \end{cases} $$

but what about for arbitrary $n$? The same question for the Chebyshev polynomial $U_n(x).$


1 Answer 1


Well, suppose it depends upon how closed you want the formula ...?

The $n$'th Chebyshev polynomial of the first kind may be written: $$ T_n(x) = \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{n}{2}(-1)^k \frac{(n-k-1)!}{k!(n-2k)!} (2x)^{n-2k}$$ so you could just integrate term by term. For $i+n$ odd the integral is zero and when even (if I didn't make a mistake in my calculation): $$ I_{i,n} =\sum_{k=0}^{\lfloor n/2 \rfloor} \frac{(-1)^k \;n \;(n-k-1)! \; 2^{n-2k+1}}{k!(n-2k)! \; 2 (n+1+i-2k)} .$$ There is a similar formula for $U_n$ (see e.g. wiki page). I suspect that there are no easy simplifications since the $T_n$'s do not have nice orthogonal behavior with respect to the Lebesgue measure. But that's just a guess.


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